Answer: the amplitude of the motion is 0.048 m
Explanation:
Given that;
mass m = 3.3 kg
spring constant k = 260 N/m
When the object is x = 0.017 m from its equilibrium position,
it is moving with a speed of v = 0.40 m/s
amplitude of the motion = ?
we know that;
formula for Potential of the string PE = 1/2kx²------1
Kinetic energy KE = 1/2mv² --------2
Now expression for the maximum potential energy of the spring is
PE_max = 1/2 kx²_max -----3
Now from conservation of Energy
PE_spring = PE + KE
from our equation 1, 2 and 3
1/2kx²_max = 1/2kx² + 1/2mv²
now we rearrange for x_max
x²_max = (1/2kx² + 1/2mv²) / 1/2k
x²_max = (kx² + mv²) / k
x_max = √ [(kx² + mv²) / k]
now we substitute our values into the equation
x_max = √ [(260(0.017)² + 3.3×(0.40)²) / 260]
= √((0.07514 + 0.528) / 260)
= √(0.60314 / 260)
= √ 0.002319
x_max = 0.048 m
Therefore the amplitude of the motion is 0.048 m
An astronaut on the Moon releases a rock from rest and allows it to drop straight downward. If the acceleration due to gravity on the Moon is 1.62m/s2 and the rock falls for 2.4s before hitting the ground, what is its speed just before it lands?
Answer:
speed before landing = 3.9 m/s (3 s.f.)
Explanation:
As rock is released from rest, u = 0 m/s a = 1.62 m/s² t = 2.4 s v = ?
v = u + at
v = 0 + (1.62 x 2.4)
v = 3.888 = 3.9 s (3 s.f.)
Hope this helps!
The speed of the rock before hitting the ground is 3.89 m/s
The given parameters;
acceleration due to gravity on moon, g = 1.62 m/s²
time taken for the object to fall, t = 2.4 s
To find:
the speed of the object before hitting the ground;The maximum height of fall of the rock is calculated as;
[tex]h = v_0t + \frac{1}{2} gt^2\\\\v_0 = 0\\\\h = \frac{1}{2} gt^2\\\\h = 0.5 \times 1.62 \times 2.4^2 \\\\h = 4.67 \ m[/tex]
The speed of the rock before hitting the ground is calculated as;
[tex]v_f^2 = v_0 ^2 + 2gh\\\\v_f^2 = 0 + 2\times 1.62 \times 4.67\\\\v_f^2 = 15.13\\\\v_f = \sqrt{15.13} \\\\v_f = 3.89 \ m/s[/tex]
Thus, the speed of the rock before hitting the ground is 3.89 m/s
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Newton's 2nd Law describes the relationship between force, mass, and acceleration of an object with the formula F=ma. Which scenario below will result in the greatest accleration? (AKS 8b / DOK 1)
A. A 20 kg rock thrown with a 100 N force.
B. A 10,000 kg rocket launched with 100,000 N of force.
C. A 2,000 kg car accelerated with a 1000 N force from the engine.
D. A 50 kg student pushed in a chair with a force of 1000 N.
Answer:D
Explanation:If you want to find the acceleration do Force/mass.
The greatest magnitude of acceleration is of 20 m/s², by the student of 50 kg pushing the chair with 1000 N of force. Hence, option (D) is correct.
As per the Newton's second law, " The applied force on an object is equal to the product of mass of object and acceleration cause by the applied force. Then, the mathematical expression is,
F = ma
Here, F is the applied force, m is the mass and a is acceleration.
(A)
For 20 kg mass and 100 N force, the acceleration is,
F = ma
100 = (20)a
a = 5 m/s²
(B)
For 10,000 kg mass of rocket and 100,000 N force, the acceleration is,
F = ma
100 ,000= (10,000)a
a = 10 m/s²
(C) For a 2,000 kg mass of car and force of 1000 N, the acceleration is,
F = ma
1000 = (2000)a
a = 0.5 m/s²
(D) For a mass of 50 kg student and applied force of 1000 N, the acceleration is,
F = ma
1000 = (50)a
a =20 m/s².
Thus, we can conclude that the greatest magnitude of acceleration is of 20 m/s², by the student of 50 kg pushing the chair with 1000 N of force. Hence, option (D) is correct.
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A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer
Answer:
After 3secondsExplanation:
A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer
Using the formula for calculating speed expressed as;
Speed = Distance/Time
Given;
Distance = 3km = 3000m
Speed = 1000m/s
Required
How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer (Time)
From the formula;
Time = Distance/speed
Time = 3000/1000
Time = 3seconds
Hence the sound of the aircraft is heard after 3 seconds
A section of a sphere has a radius of curvature of 0.72 m. If this section is painted with a reflective coating on both sides, what is the focal length of:
Answer:
The value is [tex]f = 0.36 \ m[/tex]
Explanation:
From the question we are told that
The radius of curvature is [tex]R = 0.72 \ m[/tex]
Generally the radius of curvature is mathematically represented as
[tex]R = 2f[/tex]
Here [tex]f[/tex] is the focal length hence
[tex]f = \frac{R}{2}[/tex]
=> [tex]f = \frac{0.72}{2}[/tex]
=> [tex]f = 0.36 \ m[/tex]
5. Correct the statement : Human being
can hear sounds of frequencies upto
60000Hz.
Answer:
Humans can detect sounds in a frequency range from about 20 Hz.
What will happen to the ice cubes in the two spoons? How will they be different?
Answer: the ice cube to melt more quickly on the metal block than on the wooden block.
Explanation:
What kinds of forces can act on an object
A water-skier of mass 75.0 kg initially at rest is being pulled due east by a horizontal towrope. The rope exerts a force of 365 N (east). The water (and air) exerts a combined average frictional force of 190 N (in the opposite direction). How fast will the skier be moving after a distance of 38.0 m?
Answer:
The skier will be moving at 13.31 m/s.
Explanation:
To calculate the velocity of the skier we need to find the acceleration, as follows:
[tex] \Sigma F = ma [/tex]
[tex] F_{r} - F_{f} = ma [/tex]
Where:
[tex] F_{r}[/tex]: is the force due to the rope = 365 N
[tex] F_{f}[/tex]: is the combined average frictional force = 190 N
m: is the mass = 75.0 kg
[tex] a = \frac{365 N - 190 N}{75.0 kg} = 2.33 m/s^{2} [/tex]
Now, we can calculate the velocity of the skier by using the following kinematic equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex] v_{f}[/tex]: is the final velocity =?
[tex] v_{0}[/tex]: is the initial velocity = 0 (the skier is initially at rest)
d: is the distance = 38.0 m
[tex] v_{f} = \sqrt{2*2.33 m/s^{2}*38.0 m} = 13.31 m/s [/tex]
Therefore, the skier will be moving at 13.31 m/s.
I hope it helps you!
1. On a force vs. mass graph, what would be the slope of the line?
2. On a Free Body Diagram, if the forces are all balanced, what do you know about the
object? Can it be moving?
1. By Newton's second law,
F = m a
so the slope of the line would represent the mass of the object.
2. If all the forces are balanced, then the object is in equilibrium with zero net force, which in turn means the object is not accelerating. So the object is either motionless or moving at a constant speed.
A monatomic ideal gas with volume 0.230 L is rapidly compressed, so the process can be considered adiabatic. If the gas is initially at 1.01 105 Pa and 3.00 102 K and the final temperature is 489 K, find the work done by the gas on the environment, Wenv.
Answer:
The value is [tex]W = - 17.53 \ J[/tex]
Explanation:
From the question we are told that
The volume is [tex]V = 0.230 \ L = 0.230 *10^{-3} \ m^{-3}[/tex]
The initial pressure is [tex]P_1 = 1.01105 \ Pa[/tex]
The initial temperature is [tex]T_1 = 3.00*10^2 \ K[/tex]
The final temperature is [tex]T_2 = 489 \ K[/tex]
Generally for an adiabatic process the workdone is mathematically represented as
[tex]W = - \Delta U[/tex]
Here [tex]\Delta U[/tex] is the internal energy of the system which is mathematically represented as
[tex]\Delta U = \frac{3}{2} * nR \Delta T[/tex]
So
[tex]W = - \frac{3}{2} * nR \Delta T[/tex]
Generally from ideal gas equation we have that
[tex]n = \frac{P_1V }{ RT_1 }[/tex]
Here R is the gas constant with value [tex]R = 8.314 J/mol\cdot K[/tex]
So
[tex]n = \frac{1.01 *0^{5} * 0.230 *10^{-3}}{ 8.314 * 3.0*10^2 }[/tex]
=> [tex]n = 0.009313 \ mol[/tex]
So
[tex]W = - \frac{3}{2} * 0.009313 * 8.314 * (451 - 3.00*10^2)[/tex]
=> [tex]W = - 17.53 \ J[/tex]
A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 3.50 s. Find the force constant of the spring.
Answer:
17.71N/m
Explanation:
The period of the spring is expressed according to the expression;
[tex]T = 2 \pi \sqrt{\frac{m}{k} } \\[/tex]
m is the mass of the object
k is the force constant
Given
m = 5.50kg
T = 3.50s
Substitute into the formula;
[tex]T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} = \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k = 17.71N/m[/tex]
Hence the force constant of the spring is 17.71N/m
A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. How long until the cannonball hits the ground?
96.7 s
55.3 s
65.3 s
45.5 s
Answer:
96.7 s
Explanation:
Time of flight in projectile can be calculated thus:
T = 2 × u × sin ϴ/ g
Where;
T = time of flight (s)
u = initial velocity (m/s)
ϴ = Angle of projectile (°)
g = acceleration due to gravity (9.8m/s²)
Based on the provided information; u = 670m/s, ϴ = 45°
Hence, using T = 2.u.sin ϴ/ g
T = 2 × 670 × sin 45° ÷ 9.8
T = 1340 × 0.7071 ÷ 9.8
T = 947.52 ÷ 9.8
T = 96.68
T = 96.7s
Answer5 ms 3. A football player has a mass of 95 kg, and he is running with a velocity of 15 m/s. What is his momentum? Answer:
Answer:
1425kgm/s
Explanation:
Given parameters:
Mass = 95kg
Velocity = 15m/s
Unknown:
Momentum = ?
Solution:
The momentum of a body is the amount of motion it posses;
Momentum = mass x velocity
Insert the parameters and solve;
Momentum = 95 x 15 = 1425kgm/s
Elements known as noble gases do not have the ability or the room in their valence
electron shell to form bonds with other elements.
O True
O False
Answer:
true
Explanation:
noble gases are octet meaning they they have eight electrons in their outer shell so the are stable
A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 7.1 m making 3.9 revolutions every 9.4 second. What is the centripetal acceleration of the ball?
Answer:
48,2 m/s²
Explanation:
We're gonna use the Centripetal Acceleration formula: v² / r but before that, we got to know the velocity, that is not shown clearly to us, so....
To know the velocity let's calculate the distance that the ball traveled
The circumference of a circle formula is:
2piR
2 . 3,14 . 7,1 | That is equal to 44,588 m
We know that the ball traveled this distance 3,9 times, so...
44,588 . 3,9 = 173,8932 m
Ok, now we have the distance, just need to know the time, that is 9.4 seconds.
Velocity = Distance / Time
V = 173,8932 / 9,4
V = 18,5 (approximate)
So...
We are back to the first formula:
Ca = v² / r
Ca = 18,5² / 7.1
Ca = 48,2 m/s² (approximate)
I hope it is correct, hahaha.
What happens when light from the sun passes through any type of matter
Answer:
its it filled with energy
Answer:
The light slows down.
Explanation:
help asap
what is the force responsible for seafloor spreading and the formation of new ocean floor at mid ocean ridges
Answer:
magma
Explanation:
I wanna think that that's right if it's not in so sorry but I'm pretty sure it's magma
Intrusive magma forced up towards the ocean ridge is the force responsible for seafloor spread. The correct option is B.
What is seafloor spreading?Seafloor spreading is a process that occurs at mid-ocean ridges where the oceanic crust is produced by volcanic activity and then moves away from the ridge.
Plate tectonics' continental drift is explained by seafloor spreading. When oceanic plates diverge, tensile stress causes fractures in the lithosphere.
Ridge push occurs when plates are not subducting and are driven by gravity as they slide off the elevated mid-ocean ridges.
Magma rises from the fractures at a spreading center and cools on the ocean floor to form a new seabed.
The spreading rate is the rate at which an ocean basin widens due to seafloor spreading and determines whether the ridge is fast, intermediate, or slow.
Thus, the correct option is B.
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The activity of mining has direct negative effects on
a. atmosphere
b.hydrosphere
c.biosphere
d.geosphere
Answer:
A
Explanation:
When coal is burned it releases many harmful chemicals into the air, such as methane (greenhouse gas) and contributes to global warming
A person walks 5 meters north, which is vector 1. Then he walks 4 meters east, which is vector 2. The displacement vector is vector 3. Draw the vectors and calculate displacement and the angle between vectors 1 and 3
Explanation:
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A 120V AC signal is available at the wall port. You are required to convert it to a constant DC voltage of 12V for a laptop to operate. What are the three electrical components you will be using to achieve the same? Name the components in order and the function it performs.
Answer:
The following components shall be used in given order:
1. Step Down Transformer 2. Diode as Rectifier 3. Capacitor
Explanation:
The following components shall be used in given order:
1. Step Down Transformer 2. Diode as Rectifier 3. Capacitor
1.STEP DOWN TRANSFORMER:
The step down transformer is first used to convert the 120 V AC signal into 12 V AC signal. Because the output required is of 12 V.
2.DIODE AS RECTIFIER:
Now, the PN Junction diode is used as a rectifier to convert the 12 V AC signal to 12 V DC signal.
3.CAPACITOR:
Now, the output is in waveforms. So, in order to linearize it in the form of DC current, we use capacitor. The main function of capacitor is to smoothen the signal into a straight line.
calculate the load placed 10m from the fulcrum that can be balanced by an effort of 5 N applied at a distance of 4 m from the fulcrum in a lever
Answer:
A = 2 m from fulcrum
Explanation:
Product of anti clockwise = Product of clockwise moment
5 × 4 = 10 × A
20 = 10 x A
A = 20 / 10
A = 2 m from fulcrum
A rock is at the top og a 20 meter tall hill. The rock has a mass of 10 kg. How much potential energy does it have
Answer:
1962 joules
Explanation:
m = 10 kg
h = 20 m
g = 9.81 ms^-2
PE = ?
PE = MGH
PE = 10 x 9.81 x 20
PE = 1962 joules
A rock is thrown straight up with an initial speed of 22 m/s. How long will it be in the air before it returns to the thrower?
Answer:
The time spent in air by the rock before it returns to the thrower is 4.5 s
Explanation:
Given;
initial velocity of the rock, u = 22 m/s
Neglecting air resistance, the time it takes the rock to go up equals the time it takes it to return to the thrower. This is also called time of flight.
The time to go up is calculated as;
v = u - gt
where;
v is the final velocity at the maximum height
0 = u - gt
gt = u
t = u / g
t = 22 / 9.8
t = 2.245 s
thus time to go up = 2.245 s
and time to return to the thrower = 2.245s
The total time spent in air by the rock before it returns to the thrower
= 2.245 s + 2.245s
= 4.5 s
Starting with the definitions of momentum and kinetic energy, derive an equation for the kinetic energy of a particle expressed as a function of its momentum.
Answer:
[tex]K.E = (\frac{1}{2})Pv[/tex]
Explanation:
The momentum of a particle is defined as the product of its mass and velocity:
[tex]P = mv[/tex] -------------------- equation (1)
where,
P = momentum of the particle
m = mass of the particle
v = velocity of the particle
The kinetic energy of the particle is given as follows:
[tex]K.E = (\frac{1}{2})mv^2\\\\K.E = (\frac{1}{2})v(mv)[/tex]
using equation (1), we get:
[tex]K.E = (\frac{1}{2})Pv[/tex]
A box takes 450 N to start moving when the coefficient of static friction is 0.25. What is
the box's weight?
a car is initially at rest. ten seconds later it is moving 30 m/s. what is the car's acceleration
Answer:
The car's acceleration is [tex]3\ m/s^2[/tex]
Explanation:
Constant Acceleration Motion
It's a type of motion in which the velocity of an object changes uniformly over time.
The final speed is given by:
[tex]v_f=v_o+at[/tex]
Where a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
Using the equation above we can solve for a:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
The car is initially at rest (vo=0) and t=10 seconds later it's moving at vf=30 m/s, thus its acceleration is:
[tex]\displaystyle a=\frac{30-0}{10}[/tex]
[tex]\displaystyle a=\frac{30}{10}=3[/tex]
[tex]a = 3\ m/s^2[/tex]
The car's acceleration is [tex]3\ m/s^2[/tex]
A stretched string is 1.95 m long and has a mass of 19.1 g. When the string oscillates at 440 Hz, which is the frequency of the standard A pitch, transverse waves with a wavelength of 16.7 cm travel along the string. Calculate the tension T in the string.
T= ________ N
Answer:
The tension in the string is 52.89 N.
Explanation:
Given;
mass of the string, m = 19.1 g = 0.0191 kg
length of the string, L = 1.95 m
⇒mass per unit length, μ = 0.0191 / 1.95 = 0.009795 kg/m
Also Given;
frequency of the string, F = 440 Hz
wavelength of the sound wave, λ = 16.7 cm = 0.167 m
⇒the speed of the wave, v = Fλ = 440 x 0.167 = 73.48 m/s
The tension T in the string is calculated as;
[tex]v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu}\\\\T = v^2 \mu\\\\T = (73.48)^2 (0.009795)\\\\T = 52.89 \ N[/tex]
Therefore, the tension in the string is 52.89 N.
The physical quantity which have both magnitude and direction is a vector?why?
Answer:
hiiiiiiiiiiiii baby.........
As the mass of a body increases, its gravitational force of attraction to the Earth...
Answer:
the answer may be mass and distance
A large bagel spins with angular speed w about its center. A smaller bagel spins with triple the angular speed.
How does the period Tlarge of the large bagel compare with the period Tsmall of the small bagel?
Answer:
T large = 3T small
Explanation:
It's just 3 times larger